How do we resolve the paradox in 1 D Ising model at $T=0$ where we have two ground states possible and accordingly the mean value of magnetization is $0$ but there is spontaneous symmetry breaking and the system chooses one state and the net magnetization is non zero. Is it related to the fact that when we perform a measurement on the system, the system collapses into one of the two ground states and thus we get non zero magnetisation?

But this argument also seems somewhat wrong because say all spins up and all spins down are the two eigenstates of the Hamiltonian. Then since both states have the same energy and are equally probable, any linear combination of these states will also be an eigenstate of the Hamiltonian. Then how can we say that a state which is an eigenstate (but a linear combination) will collapse into one of the eigenstates?

  • 2
    $\begingroup$ The usual way this is resolved is that small fluctuations in the ambient magnetic field break the symmetry. After all, you can never truly have zero field. $\endgroup$ Feb 25, 2018 at 20:23
  • $\begingroup$ Where does this stray mag field come from? Is it due to electrons? And how does this extra field help to resolve the paradox? $\endgroup$
    – Draco_1125
    Feb 25, 2018 at 20:24
  • $\begingroup$ Having any external field immediately causes one spin-aligned state to be preferred over another. Taking it to zero doesn't change this alignment. As to where it comes from, at some level we don't really care. $\endgroup$ Feb 25, 2018 at 20:26
  • $\begingroup$ How can you say that this stray field will homogenously pointing along one direction so that one of the spin states (either all up or all down) will be preferred? $\endgroup$
    – Draco_1125
    Feb 25, 2018 at 20:30
  • $\begingroup$ It doesn't have to be homogeneous, because the interaction between the spins is much stronger than the field. It just has to slightly shift the energies of the various ground states, because any of the partially-antialigned excited states are still way above the ground states in energy. $\endgroup$ Feb 25, 2018 at 20:33

1 Answer 1


There are two ways to solve this "paradox".

The first one is to say, like it was suggested by probably_someone in the comments, that a realistic system will never be perfectly isolated from magnetic fields. An infinitesimal external magnetic field will be enough for one of the two ground states to be favored, thus selecting one of the two ground states in the limit $T\to 0$.

The second resolution is more formal. Apparently, if the Hamiltonian $\mathcal H(\sigma_1, \dots, \sigma_N) = \mathcal H (\{\sigma_i\})$ is invariant under reversal of all the spins $\sigma_i \to -\sigma_i$, then also the canonical probability of the state $\{\sigma_i\}$

$$P[\mathcal H\{\sigma_i\}]=\frac{e^{-\beta \mathcal H(\{\sigma_i\})}}{Z} \tag{1}$$

will be invariant under the same transformation. This means that the magnetization

$$m \equiv\langle \sigma_i \rangle = \sum_{\{\sigma_i\}} \sigma_i P[\mathcal H (\{\sigma_i\})]$$

must always be $0$, since $m$ and $-m$ occur with equal probability.

For a finite volume system, the correctness of this argument is irrefutable: there are no (true) phase transitions at finite volume. A finite-volume Ising model will only apparently select a defined magnetization as $T\to0$, but if you wait long enough (an exponentially long time with the system size $N$), the magnetization will eventually reverse, over and over.

However, as $V \to \infty$, $(1)$ becomes only formal, since the partition function diverges. In this case, this argument is incorrect, and it can be shown that while

$$\lim_{V\to \infty} \lim_{h \to 0^{\pm}} m = 0$$

you have

$$ \lim_{h \to 0^{\pm}} \lim_{V\to \infty} m = \pm 1$$

This is what we call spontaneous symmetry breaking: even in the absence of an external field (which would explicitly break the symmetry), a single ground state is selected over the two possible ground states.


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